Assignment 3
Prithviraj Kadiyala
October 29, 2018
1 Questions{16/16}
1.1 Question 1
1.1.1 a
pexp(120, 1/95, lower.tail = FALSE)## [1] 0.28275971.1.2 b
PH = read.csv("PHISHING.csv")
hist(PH$INTTIME,main = "Phishing ")
hist(rexp(267, rate = 1/95))
The two graphs are very similar to each other.
Yes, the data appears to follow an exponential distribution.
1.2 Question 2
1.2.1 a
alpha=3
beta=0.07
mean1=alpha*beta
mean1## [1] 0.21var1=alpha*beta^2
var1## [1] 0.01471.2.2 b
alpha=3
beta=0.07
mean1=alpha*beta
var1=alpha*beta^2
lower=mean1-3*var1^0.5
upper=mean1+3*var1^0.5
paste("Interval is","(",lower,",",upper,")")## [1] "Interval is ( -0.153730669589464 , 0.573730669589464 )"1.3 Question 3
1.3.1 a
alpha1=2
beta1=2
mean1=alpha1*beta1
var1=alpha1*beta1^2
mean1## [1] 4var1## [1] 8alpha2=1
beta2=4
mean2=alpha2*beta2
var2=alpha2*beta2^2
mean2## [1] 4var2## [1] 161.3.2 b
#Formula A
p1=pgamma(1,shape=2,scale = 2)
p1## [1] 0.09020401#Formula B
p2=pgamma(1,shape=1,scale = 4)
p2## [1] 0.2211992Therefore, Formula B has higher probability of generating a human reaction in less than 1 minute.
1.4 Question 4
1.4.1 a
pweibull(2,shape = 4, scale = 2)## [1] 0.63212061.4.2 b
alpha=2
beta=4
mean1=beta^(1/alpha)*gamma((alpha+1)/alpha)
var1=beta^(2/alpha)*(gamma((alpha+2)/alpha)- (gamma((alpha+1)/alpha))^2)
sd1=sqrt(var1)
mean1## [1] 1.772454sd1## [1] 0.92650281.4.3 c
alpha=2
beta=4
mean1=beta^(1/alpha)*gamma((alpha+1)/alpha)
var1=beta^(2/alpha)*(gamma((alpha+2)/alpha)- (gamma((alpha+1)/alpha))^2)
sd1=sqrt(var1)
lower=mean1-2*sd1
upper=mean1+2*sd1
paste("Interval is","(",lower,",",upper,")")## [1] "Interval is ( -0.0805516497989005 , 3.62545935160993 )"1.4.4 d
1-pweibull(6,shape = 4, scale = 2)## [1] 0No.
1.5 Question 5
1.5.1 a
alpha=2
beta=9
mean1=alpha/(alpha+beta)
var1=(beta*alpha)/((alpha+beta)^2*(alpha+beta+1))
mean1## [1] 0.1818182var1## [1] 0.01239669Mean = 0.18182 Variance = 0.01240
1.5.2 b
1-pbeta(0.4,shape1 = 2, shape2 = 9)## [1] 0.04635741.5.3 c
pbeta(0.1,shape1 = 2, shape2 = 9)## [1] 0.26390111.6 Question 6
1.6.1 a
alpha = 2
beta = 161.6.2 b
alpha=2
beta=16
mean=beta^(1/alpha)*gamma((alpha+1)/alpha)
var=beta^(2/alpha)*(gamma((alpha+2)/alpha)-(gamma((alpha+1)/alpha))^2)
mean## [1] 3.544908var## [1] 3.4336291.6.3 c
1-pweibull(6,shape=2,scale=16^(1/2))## [1] 0.10539921.7 Question 7
1.7.1 a
P(x,y)=\(\frac{1}{36}\)
1.7.2 b
\[ P_1(x) = \frac{1}{6}\\ P_2(y) = \frac{1}{6} \]
1.7.3 c
\[ \begin{align} P(x|y) &= \frac{p(x,y)}{p(y)}=\frac{\frac{1}{36}}{\frac{1}{6}}=\frac{1}{6} \end{align} \]
\[ \begin{align} P(y|x) &= \frac{p(x,y)}{p(x)}=\frac{\frac{1}{36}}{\frac{1}{6}}=\frac{1}{6} \end{align} \]
1.7.4 d
Since the conditional probabilities and the unconditional probabilities are equivalent, x and y don't affect each other.1.8 Question 8
1.8.1 8a
8a
1.8.2 8b
8b
1.9 Question 9
1.9.1 a
\[ f(x,y)=\frac{e^{-y/10}}{10y},\;0<y<x<2y \]
For marginal density \(f(y)\)
\[ \begin{align} f(y)&=\int^\infty_{-\infty}f(x,y)dx \\ &= \int^{2y}_y\left[\frac{e^{-y/10}}{10y}\right]dx \\ &=\left[\frac{1}{10}\frac{e^{-y/10}}{y}\biggr\rvert^{2y}_y\right] \\ &= \frac{1}{10}e^{-y/10} \end{align} \]
1.9.2 b
to Find E(y) \[ \begin{align} E(y)&=\int_y y.f(y)dy\\ &=\int^\infty_0 \left[\frac{1}{10}e^{\frac{-y}{10}}\right]ydy\\ &=\left[\left[-(10+y)\right]\right]^\infty_0 \\ &= 10 \end{align} \]
1.10 Question 10
1.10.1 10a
10a
1.10.2 10b
